Leetcode 15 - 3 sum, using binary search, TLE error, but great workout using Array.BinarySearch method.

Leetcode15_3sum_binarySearch.cs

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;

namespace _16_3Sum
{
    /*
     * Leetcode 15: 3 sum
     * https://leetcode.com/problems/3sum/
     * 
     * Review the algorithm on code review of stackexchange.com
     * http://codereview.stackexchange.com/questions/37922/given-an-array-find-any-three-numbers-which-sum-to-zero?rq=1
     */
    class Program
    {
        static void Main(string[] args)
        {
            ThreeSumTestCase(); 
        }

        private static void ThreeSumTestCase()
        {
            // test 3 sum            
            // 2 lists, one is -1, 0, 1, second one is -1, -1, 2
            int[] array = new int[6] { -1, 0, 1, 2, -1, -4 };

            IList<IList<int>> triplets = ThreeSum(array);

            Debug.Assert(triplets.Count == 2);
            Debug.Assert(String.Join(",", triplets[0].ToArray()).CompareTo("-1,-1,2") == 0);
            Debug.Assert(String.Join(",", triplets[1].ToArray()).CompareTo("-1,0,1") == 0);
        } //

        /*
         * Dec. 26, 2016
         * http://codereview.stackexchange.com/questions/37922/given-an-array-find-any-three-numbers-which-sum-to-zero?rq=1
         * 
         * May 17, 2016
         * Work on this 3 sum close algorithm
         * 
         * Given an array S of n integers, are there elements a, b, c in S 
         * such that a + b + c = 0? Find all unique triplets in the array 
         * which gives the sum of zero.
           Note:
            Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
            The solution set must not contain duplicate triplets.  
         * 
            For example, given array S = {-1 0 1 2 -1 -4},
            A solution set is:
            (-1, 0, 1)
            (-1, -1, 2)
         * 
         * TLE - time limited exceeded 
         * O(n*n*logn) 
         * 
         * Best time complexity using two points - O(n*n)
         * 
         * Warning: Do not use 
         * int[] searchArray = nums.Skip(j + 1).ToArray(); 
         * The above statement will take O(n) time, n is size of the array
         * 
         * 
         */
        public static IList<IList<int>> ThreeSum(int[] nums)
        {            
            IList<IList<int>> results = new List<IList<int>>();            

            if (nums == null || nums.Length == 0)
                return results;

            Array.Sort(nums);

            int len = nums.Length;
            HashSet<string> foundNos = new HashSet<string>(); 
            for (int i = 0; i < len - 2; i++)  // len = 3, test case passes! 
            {
                for(int j = i+1; j < len; j++)
                {
                    //int[] searchArray = nums.Skip(j + 1).ToArray();  // O(n) -> make algorithm O(n*n*n)
                    int searchValue   = -1 * (nums[i] + nums[j]);
                    int index = Array.BinarySearch(nums, j+1, len-j-1, searchValue);  // O(logn)

                    if (index >= 0)
                    {
                        int[] numberAsc = new int[] { nums[i], nums[j], nums[index] };
                        string key = PrepareKey(numberAsc, ',');

                        if (foundNos.Contains(key))
                            continue;

                        foundNos.Add(key);

                        IList<int> threeNos = numberAsc.ToList(); 
                        results.Add(threeNos);                        
                    }
                }
            }

            return results;
        }      
        
        private static string PrepareKey(int[] numbers, char delimiter)
        {
            if(numbers == null || numbers.Length == 0)
                return "";

            string key = string.Empty;
            for (int i = 0; i < numbers.Length; i++ )
            {
                key += numbers[i] + delimiter.ToString();
            }

            return key; 
        }
    }
}